Question

The bach emf in the inductance coil is 200 V. when the current in the coil changes from zero to 2A in 0.01 sec. calculate the self inductance of the coil.​

Answer 1

Answer:

1H

Explanation:

e=L×di/dt

200=L×2-0/0.01

L=1H

Answer 2

Answer:

The self inductance of coil is 1 Henry .

Explanation:

Given as :

The back emf of inductance coil = emf = 200 v

The current changes from 0 to 2 Amp

The rate of change of current = 0.01 sec

Let the self inductance of coil = L Henry

According to question

back emf of inductance coil = self inductance × $\dfrac{\partial i}{\partial t}$

Or, emf = L × $\dfrac{\Delta I}{\Delta T}$

Or, 200 = L ×  $\dfrac{2-0}{0.01}$

Or, 200 = L × 200

∴   L = $\dfrac{200}{200}$

i.e L = 1 Henry

So, The inductance of coil = L = 1 Henry

Hence, The self inductance of coil is 1 Henry . Answer